The Integral is Gold.

In General to prove: $\displaystyle\int_{a}^{b} \dfrac{f(x)}{f(a + b - x) + f(x)} dx = \dfrac{b - a}{2}$

Let $I = \displaystyle\int_{a}^{b} \dfrac{f(x)}{f(a + b - x) + f(x)} dx$ and $u = a + b - x$

$\implies I = \displaystyle\int_{a}^{b} \dfrac{f(a + b - u)}{f(u) + f(a + b - u)} du$

Since $u$ is a dummy variable we can replace $u$ by $x$ to obtain:

$I = \displaystyle\int_{a}^{b} \dfrac{f(a + b - x)}{f(x) + f(a + b - x)} dx$

Adding $I = \displaystyle\int_{a}^{b} \dfrac{f(x)}{f(a + b - x) + f(x)} dx$ and $I = \displaystyle\int_{a}^{b} \dfrac{f(a + b - x)}{f(x) + f(a + b - x)} dx$

$\implies 2I = \displaystyle\int_{a}^{b} dx \implies \boxed{I = \dfrac{b - a}{2}}$ .

Using the above result $\implies \displaystyle\int_{-1}^{b} \dfrac{f(x)}{f(-1 + b - x) + f(x)} dx =$ $\dfrac{b + 1}{2} = \dfrac{b^2}{2}$

$\implies b^2 - b - 1 = 0$

Using the positive root we obtain $b = \dfrac{1 + \sqrt{5}}{2} = \phi \approx \boxed{1.61803}$

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$\therefore$ The desired integral is: $\displaystyle\int_{-1}^{\phi} \dfrac{f(x)}{f(\phi - 1 - x) + f(x)} dx = \dfrac{\phi + 1}{2} = \dfrac{\dfrac{1 + \sqrt{5}}{2} + 1}{2} = \dfrac{\dfrac{3 + \sqrt{5}}{2}}{2} = \dfrac{\phi^2}{2}$ .

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As an example using $f(x) = x^2$ and $G(x) = \dfrac{x^2}{(\phi - 1 - x)^2 + x^2}$ the graph below shows the area of the region bounded by $G(x)$ on $[-1,\phi]$ .